Question: Is ${599592}$ divisible by $9$ ?
A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {599592}= &&{5}\cdot100000+ \\&&{9}\cdot10000+ \\&&{9}\cdot1000+ \\&&{5}\cdot100+ \\&&{9}\cdot10+ \\&&{2}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {599592}= &&{5}(99999+1)+ \\&&{9}(9999+1)+ \\&&{9}(999+1)+ \\&&{5}(99+1)+ \\&&{9}(9+1)+ \\&&{2} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {599592}= &&\gray{5\cdot99999}+ \\&&\gray{9\cdot9999}+ \\&&\gray{9\cdot999}+ \\&&\gray{5\cdot99}+ \\&&\gray{9\cdot9}+ \\&& {5}+{9}+{9}+{5}+{9}+{2} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${599592}$ is divisible by $9$ if ${ 5}+{9}+{9}+{5}+{9}+{2}$ is divisible by $9$ Add the digits of ${599592}$ $ {5}+{9}+{9}+{5}+{9}+{2} = {39} $ If ${39}$ is divisible by $9$ , then ${599592}$ must also be divisible by $9$ ${39}$ is not divisible by $9$, therefore ${599592}$ must not be divisible by $9$.